Conversion from continued fraction of binomial expansion

to that of power function

We start from the last result

  in special form

mentioned yesterday. Its reciprocal is just

   in special form

where we set z=(1+x)^y to 1/z.

Now we use the following transformation

Setting to the expression of , we get

   in special form

Applying Rule 2(down->front and next) of signs as in 0109, we get

   in special form

Finally, setting x=x-1 to make the term of x^y, we get

Thus

  in special form